How to evaluate $\int_{0}^{1}\arcsin\left ( x\arccos x \right )\mathrm{d}x$ How to evaluate the following integral

$$\mathcal{I}=\int_{0}^{1}\arcsin\left ( x\arccos x \right )\mathrm{d}x$$

Mathematica can only give me a numerical solution $\mathcal{I}\approx 0.40926\cdots$.
So I want to know if there is a closed form for the integral.
 A: By using the formula,
$$\int_a^b f(x)\space dx=(b-a)\sum_{n=1}^\infty\sum_{m=1}^{2^n-1}\frac{(-1)^{m+1}}{2^{n}}f\left(a+\frac{m(b-a)}{2^{n}}\right)$$
We have
$$\int_0^1\sin^{-1}(x\cos^{-1}x)~dx=\sum\limits_{n=1}^\infty\sum\limits_{m=1}^{2^n-1}\dfrac{(-1)^{m+1}}{2^n}\sin^{-1}\left(\dfrac{m}{2^n}\cos^{-1}\dfrac{m}{2^n}\right)$$
A: The most we can do is find a closed form expression in elementary functions with a given precision.
First, let's note that
$$x\arccos x < 1$$
$$0 \leqslant  x\leqslant 1$$
Taking into account this equality we use Taylor expansion of $\arcsin x$
$$\arcsin x =x + \frac{x^3}{6}+ \frac{3}{40}x^5 + ...$$
$$x^2<1$$
Considering only the first three terms here we need to compute three integrals
$$\int_{0}^{1} x\arccos x \; dx=\frac{\pi}{8}$$
$$\int_{0}^{1}\left ( x\arccos x \right )^3\mathrm{d}x=\frac{3\pi(2\pi^2-15)}{512}$$
$$\int_{0}^{1}\left ( x\arccos x \right )^5\mathrm{d}x=\frac{5\pi(54\pi^4-1470\pi^2+9485)}{16588}$$
Using the obtained results we get
$$\mathcal{I}\approx \frac{5\pi(54\pi^4-606\pi^2+58301)}{442368}$$
The approximation error here is about $0.00034$
If we want a more precise expression, we have to include the next term of the Taylor series, $\frac{5}{112}x^7$, etc.
A physicist's approach
I would prefer the following approach
$$\mathcal{I}=\int_{0}^{1}\arcsin\left ( x\arccos x \right )\mathrm{d}x$$
To get the approximate value of $\mathcal{I}$, i push the integral operator inside the $\arcsin$ function
$$\mathcal{I}\approx \arcsin \left ( \int_{0}^{1} x\arccos x \; dx \right )=\arcsin\left ( \frac{\pi}{8} \right )=0.4036$$
The approximation error here is about $0.006$
We see that, unlike the first, the second method stands out for its simplicity and comprehensibility, although the majority of mathematicians would consider it illegal.
